ON SOME GEOMETRIC RELATIONS OF A TRIANGLE

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1 N SME GEMETRI RELTINS F TRINGLE ISMIL M ISEV, YURI N MLTSEV, ND NN S MNSTYREV bstract Fo triangle we consider the circles passing throug vertex of the triangle tangent to the oppisite side as well as to the circumcircle We prove that + r b + r c R + r, where ra, r b, r c are the radii of the these three circles, R, re the circumradius the inradius of the triangle, respectively This equation generalizes the main result from [] The main results Let R r be the circumradius the inradius of an arbitrary triangle Denote the center of the circumcircle of by onsider the circle tangent to to the circumcircle of at the vertex We denote its center by Denote by the radius of this circle ( ) nalogously, we define r b r c We will also use the stard notation for lengths angles of the triangle: c, b, a, α, β, γ, p a + b + c It is not hard to see that ( ) [, Lemma ] a cos () Fig In [], the following inequalities have been proved R + r b + r c r So the authors get a new interitation for the well-known Euler s inequality R r In this paper, we prove the following Theorem + r b + r c R + r

2 ISMIL M ISEV, YURI N MLTSEV, ND NN S MNSTYREV efore proving the theorem, let us conside supplementary lemma Lemma Suppose 0 is the midpoint of the arc not containing the point The points 0 0 are defined analogously (see Fig ) Then the area of the hexagon is equal to pr Proof First, we note that S S 0 + S 0 + S Fig However S 0 ar, S 0 br S 0 cr Thus S pr Now we can prove Theorem Proof of Theorem Let ω be the circumcircle of the triangle, ω be the tangent circle at to the circumcircle of the triangle to the side, be the tangent point of ω the side (see Fig 3) 0 Fig 3 onsider the homothetic transformation with centet such that ω is mapped to ω Then the line is mapped to the tangent of ω parallel to Therefore the point is mapped to 0, where 0 is the midpoint of the circularc not containing the point In particular, by the rchimedean lemma, is a bisector of the angle Hence we have R 0 S 0 S S 0 S S 0 S

3 N SME GEMETRI RELTINS F TRINGLE 3 y Lemma we get R It implies that + R r b r b + R r c r c S 0 + S 0 + S 0 S + r b + r c R + r pr S S Using Theorem, one can obtain a new proof of inequalities from []: orollary 3 Fony triangle the following inequalities hold: R + r b + r c r Moreover, we have equality if only if the triangle is equilateral Proof Since R r, we have + r b + r c r + R r + r r + + r b r c R + R/ R It is well-known that R r if only if the triangle is equilateral (see []) So we have equalities if only if the triangle is equilateral Theorem Fony triangle the following relation holds Proof Since Hence, + + p + r + R + Rr a br b cr c R a cos ( a ), we have ( ) cos + cos() + + (3 + cos() + cos(α β) + cos(α γ)) a br b cr c (3 + (cos β cos γ + cos α cos β + cos α cos γ) + (sin β sin γ + sin α sin β + sin α sin γ)) It is known that cos β cos γ + cos α cos β + cos α cos γ p + r R R, sin β sin γ + sin α sin β + sin α sin γ p + r + Rr R

4 ISMIL M ISEV, YURI N MLTSEV, ND NN S MNSTYREV (see []) It implies a + br b + cr c (3 + p + r R R + p + r ) + Rr R ( R 8R + p + r R + Rr ) ( R R + p + r + Rr ) This completes the proof of Theorem orollary 5 Fony triangle the following inequalities hold: + + 3R a br b cr c 3r Moreover, we have equality if only if the triangle is equilateral Proof In [], it is proved that 6Rr 5r p R +Rr+3r 3 3r p 3 3 R Hence, + + a br b cr c R (R + p + r + Rr) R + 7 ( ) R R + + R R r 3 3r R r 3r + + a br b cr c R (R + p + r + Rr) R + (6Rr 5r ) + r + Rr r R 3 3 R R + 8Rr r rr 3 3 R + 9Rr r 3 rr R because the last inequality is equivalent to R 3Rr r 0, ie (R r)(r+r) 0 (see the Euler s inequality R r) So + + a br b cr c 3R Moreover, we have the equality if only if the triangle is equilateral In the triangle let, h b, h c be the lengths of the altitudes from the vertices,,, respectively The following theorem holds Theorem 6 Fony triangle the following relation + p + Rr + r r b r c R holds Proof Since ( ), we have a cos ( ) ( ) a cos S cos ( ) cos + cos() S

5 REFERENES 5 It implies 3 + cos() + cos(β α) + cos(α γ) r b r c 3 + (cos β cos γ + cos β cos α + cos α cos γ) + (sin β sin γ + sin β sin α + sin α sin γ) 3+ p + r R R + p + r + Rr R 3+ p + r + Rr R R + p + r + Rr R This completes the proof of Theorem 6 orollary 7 Fony triangle the following inequalities hold: 9Rr r + R 3Rr + r + r b r c R 7 Moreover, we have equality if only if the triangle is equilateral Proof It is known that 6Rr 5r p R + Rr + 3r (see []) Hence + p + r + Rr r b r c R + (R + Rr + 3r ) + Rr + r 3Rr + r R + R + (6Rr 5r ) + Rr + r 9Rr r r b r c R + R Since r R, we have + 3 r ( r ) r b r c R R References [] D ndrica D Ş Marinescu New inteolation inequalities to Euler s R r Forum Geom, 7:9 56, 07 issn: [] V P Soltan S I Meĭdman Tozhdestva i neravenstva v treugol nike (Identities inequalities in a triangle) Shtiintsa, Kishinev, 98 ltai State Pedagogical University, arnaul, Russia address: isaev@uni-altairu ltai State Pedagogical University, arnaul, Russia address: maltsevyn@gmailcom ltai State University, arnaul, Russia address: akuzmina@yexru

Chapter 8. Feuerbach s theorem. 8.1 Distance between the circumcenter and orthocenter

Chapter 8. Feuerbach s theorem. 8.1 Distance between the circumcenter and orthocenter hapter 8 Feuerbach s theorem 8.1 Distance between the circumcenter and orthocenter Y F E Z H N X D Proposition 8.1. H = R 1 8 cosαcos β cosγ). Proof. n triangle H, = R, H = R cosα, and H = β γ. y the law